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SUMMATION

by hpgmat » Tue Nov 17, 2009 11:38 am
if each term in the sum a1+a2+........a n is either 7 or 77 , and the sum equals 350, which could be equal to n?

38
39
40
41
42

My Approach:

7 H + 77( N-H) = 350

7 ( H + 11 (N-H))= 350
7 ( 11 N -10 H) = 350
11N - 10H = 50 ................?????? STUCK HERE
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by palvarez » Tue Nov 17, 2009 11:39 am
n = (50+10h)/11 = 4 +h + [(6-h)/11]

h = 6 (mod 11)
h = 6 or 17, 28, 39 etc

h = 39, then n = 43-3 = 40
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by hpgmat » Tue Nov 17, 2009 11:57 am
palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]

h = 6 (mod 11)
h = 6 or 17, 28, 39 etc

h = 39, then n = 43-3 = 40
IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].

PLEASE EXPLAIN
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by palvarez » Tue Nov 17, 2009 12:01 pm
hpgmat wrote:
palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]

h = 6 (mod 11)
h = 6 or 17, 28, 39 etc

h = 39, then n = 43-3 = 40
IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].

PLEASE EXPLAIN

50+10h = 11(4+h) + (6-h)

Divide by 11.
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by hpgmat » Tue Nov 17, 2009 12:16 pm
palvarez wrote:
hpgmat wrote:
palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]

h = 6 (mod 11)
h = 6 or 17, 28, 39 etc

h = 39, then n = 43-3 = 40
IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].

PLEASE EXPLAIN

50+10h = 11(4+h) + (6-h)

I GET THAT NOW , THANKS . BUT , HOW DO YOU GET THOSE NUMBERS FOR H?
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by palvarez » Tue Nov 17, 2009 5:23 pm
hpgmat wrote:
palvarez wrote:
hpgmat wrote:
palvarez wrote:n = (50+10h)/11 = 4 +h + [(6-h)/11]

h = 6 (mod 11)
h = 6 or 17, 28, 39 etc

h = 39, then n = 43-3 = 40
IM LOST ON HOW YOU GOT FROM n = (50+10h)/11 TO 4 +h + [(6-h)/11].

PLEASE EXPLAIN

50+10h = 11(4+h) + (6-h)

I GET THAT NOW , THANKS . BUT , HOW DO YOU GET THOSE NUMBERS FOR H?
Here is the logic:

n = (50+10h)/11 = 4+h + ((6-h))/11)

if n is an integer, (6-h)/11 has to be an integer.

or 6-h is a multiple of 11
or h = 11k+6, for any integer k.

Based on the answer choices pick some number k and match it
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